How do you find the x coordinates of all points of inflection, final all discontinuities, and find the open intervals of concavity for #y=-(-3x+3)^(1/2)#?

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Answer 1 · 2017-09-29T12:31:03

The function is convex on the interval #(-oo,1)#

Calculate the first and second derivatives

#y=-sqrt(3-3x)#

The domain of definition is #I= (-oo,1]#

#dy/dx =-1/2*(-3 )(3-3x)^(-1/2)=3/2(3-3x)^(-1/2) #

#(d^2y)/dx^2=-3/2*-3/2(3-3x)^(-3/2)=9/4(3-3x)^(-3/2)#

#AA x in I#, #(d^2y)/dx^2>0#, the function is convex

graph{-sqrt(3-3x) [-16.02, 16.01, -8.01, 8.01]}